Understanding Cylinder Formation from Rectangular Sheets

When a hollow right circular cylinder is formed by joining a pair of parallel sides of a rectangular sheet, one dimension of the rectangle becomes the circumference of the cylinder's base, and the other dimension becomes the height of the cylinder. Let the dimensions of the rectangular sheet be length $L$ and width $W$.

• If we join the sides of length $L$, the circumference is $L$, and the height is $W$.
• If we join the sides of length $W$, the circumference is $W$, and the height is $L$.

The circumference of a circle is given by the formula $C = 2\pi r$, where $r$ is the radius of the base. The volume of a right circular cylinder is given by $V = \pi r^2 h$, where $h$ is the height.

Analyzing the First Rectangular Sheet: 2π × 4π

The first sheet has dimensions $2\pi$ and $4\pi$. We can form a cylinder in two ways:

Case 1.1: Joining sides of length 2π

In this case, the circumference $C = 2\pi$ and the height $h = 4\pi$.

Using $C = 2\pi r$, we have $2\pi = 2\pi r$, which gives the radius $r = 1$.

The volume of the cylinder is $V = \pi r^2 h$.

Substituting the values, $V_1 = \pi (1)^2 (4\pi) = \pi (1) (4\pi) = 4\pi^2$.

Case 1.2: Joining sides of length 4π

In this case, the circumference $C = 4\pi$ and the height $h = 2\pi$.

Using $C = 2\pi r$, we have $4\pi = 2\pi r$. Dividing both sides by $2\pi$, we get the radius $r = 2$.

The volume of the cylinder is $V = \pi r^2 h$.

Substituting the values, $V_2 = \pi (2)^2 (2\pi) = \pi (4) (2\pi) = 8\pi^2$.

Analyzing the Second Rectangular Sheet: π × 5π

The second sheet has dimensions $\pi$ and $5\pi$. We can form a cylinder in two ways:

Case 2.1: Joining sides of length π

In this case, the circumference $C = \pi$ and the height $h = 5\pi$.

Using $C = 2\pi r$, we have $\pi = 2\pi r$. Dividing both sides by $2\pi$, we get the radius $r = \frac{\pi}{2\pi} = \frac{1}{2}$.

The volume of the cylinder is $V = \pi r^2 h$.

Substituting the values, $V_3 = \pi \left(\frac{1}{2}\right)^2 (5\pi) = \pi \left(\frac{1}{4}\right) (5\pi) = \frac{5}{4}\pi^2 = 1.25\pi^2$.

Case 2.2: Joining sides of length 5π

In this case, the circumference $C = 5\pi$ and the height $h = \pi$.

Using $C = 2\pi r$, we have $5\pi = 2\pi r$. Dividing both sides by $2\pi$, we get the radius $r = \frac{5\pi}{2\pi} = \frac{5}{2}$.

The volume of the cylinder is $V = \pi r^2 h$.

Substituting the values, $V_4 = \pi \left(\frac{5}{2}\right)^2 (\pi) = \pi \left(\frac{25}{4}\right) (\pi) = \frac{25}{4}\pi^2 = 6.25\pi^2$.

Comparing Possible Cylinder Volumes

We have calculated four possible volumes from the two rectangular sheets:

• From the 2π × 4π sheet: $4\pi^2$ and $8\pi^2$.
• From the π × 5π sheet: $1.25\pi^2$ and $6.25\pi^2$.

Let's compare these values:

$4\pi^2 \approx 4 \times (3.14)^2 \approx 4 \times 9.86 = 39.44$

$8\pi^2 \approx 8 \times (3.14)^2 \approx 8 \times 9.86 = 78.88$

$1.25\pi^2 \approx 1.25 \times (3.14)^2 \approx 1.25 \times 9.86 = 12.325$

$6.25\pi^2 \approx 6.25 \times (3.14)^2 \approx 6.25 \times 9.86 = 61.625$

Comparing the exact values: $4\pi^2$, $8\pi^2$, $1.25\pi^2$, $6.25\pi^2$.

The maximum value among these is $8\pi^2$.

Summary of Cylinder Volumes

The maximum possible volume of the cylinder that can be formed is $8\pi^2$.

Revision Table - Cylinder Volume Calculation

Additional Information - Geometry Concepts

Right Circular Cylinder: A three-dimensional solid with two parallel circular bases connected by a curved surface. The axis passing through the center of the bases is perpendicular to the bases.

Circumference of a Circle: The distance around the circle. Formula: $C = 2\pi r$ or $C = \pi d$, where $d$ is the diameter.

Area of a Rectangle: Length × Width.

Forming 3D Shapes from 2D Nets: Many 3D shapes, like cylinders, cones, and cubes, can be formed by folding 2D shapes (nets). A cylinder's net is typically a rectangle and two circles for the bases. In this problem, we consider forming a hollow cylinder, which only requires the rectangular part of the net.

Understanding how the dimensions of the 2D net relate to the dimensions of the 3D shape is crucial for solving such geometry problems.